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Homework Help: Kinematic formula problem

  1. Oct 4, 2009 #1
    1. The problem statement, all variables and given/known data

    A car starts from rest and accelerates uniformly at 3 m/s2. A second car starts from rest 6 s later at the same point and accelerates uniformly at 5 m/s2. How long does it take the second car to overtake the first car?
    How would I solve this? If someone can break it down step by step, it would be great.

    2. Relevant equations
    x=x0 + v0t + 1/2at2

    3. The attempt at a solution

    I tried solving it by using the above formula and solving for t. Didnt quite work out.

    car 1's time = car two's time +6 and then solving for t.
  2. jcsd
  3. Oct 4, 2009 #2
    Find the position of the first car. Then, use that distance and set it equal to the position equation of the second car to find the time.
  4. Oct 4, 2009 #3
    I dont understand what you are saying, can you clarify please? How would I find the distance with the information given?
  5. Oct 4, 2009 #4
    Find the velocity of the first
  6. Oct 4, 2009 #5
    find the position of the first car in relation to what? the initial 6 second lead?
  7. Oct 4, 2009 #6
    Correct, the position with the 6 second lead
  8. Oct 4, 2009 #7
    (3/2)t2= (5/2)(t- 6)2

    I've tried solving for this, dosnt seem to match that answer given at the back of the book.
    Am I doing something wrong?
  9. Oct 4, 2009 #8
    What's the answer? Where are you getting those numbers?
  10. Oct 4, 2009 #9

    t2=(t1- 6) ...t1 is the time for the first car, then t2 left 6 seconds after the car, so it had ( t1- 6 ) seconds to get to the same position (at which point it would overtake car1)

    x = x0 + v0t + 1/2at12
    x = 0 + 0 + 1/2(3)t12
    x = (3/2)t12

    and for car two it is..

    x = 0 + 0 + 1/2(5)( t1 - 6 )2
    x = (5/2)( t1 - 6 )2

    so that means that x would be the same distance, given these times

    (3/2)t12 = (5/2)( t1 - 6 )2

    The answer at the back of the book is 22.7 s
    That is not the answer I get when I solve for this.
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